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EXPAND sqrt{/frac{1x}{1+x}}



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\[\Huge \sqrt{\frac{1x}{1+x}}\]

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up to power x^2

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i get \[1x+\frac{x^2}{2}  \frac{x^3}{8}  \frac{3x^4}{64}\]

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@saifoo.khan

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wolfram got \[\frac{3 x^5}{8}+\frac{3 x^4}{8}\frac{x^3}{2}+\frac{x^2}{2}x+1\]

saifoo.khan
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Unfortunately i forgot the formula! D:

wio
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Is it a maclaurin series?

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you can try that

wio
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I want to know what is meant by Expand, since that's the only thing that comes to mind.

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you can use this
\[(x+a)^n=\sum ^{n}_{ k=0} \left(\begin{matrix}n \\ k\end{matrix}\right) x^k a^{(nk)}\]

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or
\[(1+x)^n = 1+nx+\frac{(n)(n1)}{2!}(x^2)+...\]

wio
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That works for \(n\in \mathbb{N}\).

wio
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In this case you have a square root...

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still can, watch

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\[\sqrt{\frac{1x}{1+x}} \\ \\ \frac{\sqrt{1x}}{\sqrt{1+x}} \\ \\ (1x)^\frac{1}{2}(1+x)^{\frac{1}{2}}\]

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expand separately

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mix them by multiplying together

wio
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Yeah, my problem is with \(1/2 \notin \mathbb{N}\)

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I don't think it should be natural number

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i mean real number will do

wio
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How can you have summations or binomial coefficients with rational numbers?

amoodarya
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interval of x is ?

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the formula list says can use

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we use the second one (1+x)^n


amoodarya
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dw:1359798673882:dw

amoodarya
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dw:1359798819059:dw

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or maybe its \[\frac{3 x^5}{8}+\frac{3 x^4}{8}\frac{x^3}{2}+\frac{x^2}{2}x+1\]

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hmm

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based from the formula as long as its in the from ("1"+x)^n should be fine to expand, i tried it

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from \[(1x)^{1/2} \\ \\ n=1/2, x=x(for formula) \\ \\ [1\frac{x}{2}+\frac{(\frac{1}{2})(\frac{1}{2}1)}{2}(x^2)] \\ \\ 1\frac{x}{2}\frac{x^2}{8}\]